$$ \int \sin^5x \cos x\,dx $$
I used a trig substitution using $\sin$ and $\cos$, while I got two different answers, can someone explain any possible reason for the inconsistency? I got: $$ \frac{1}{6}\sin^6x $$ and $$ -\frac{1}{2}\cos^2x - \frac{1}{6}\cos^6x -\frac{1}{2}\cos^4x $$
The answer is $$\frac16\sin^6x+C.$$ But that equals $$\frac16(1-\cos^2x)^3+C=-\frac12\cos^2x+\frac12\cos^4x-\frac16\cos^6x+C'$$ which I suspect is what you intended to write,